Question

If $x=3+\sqrt{8}$, find the value of $x^2+\frac{1}{x^2}$

Answer 1

$x=3+\sqrt{3}$ $x^2=(3+\sqrt{8}{)}^2=9+8+6\sqrt{8}=17+6\sqrt{8}$

$x^2+\frac{1}{x^2}=(3+\sqrt{8}{)}^2+\frac{1}{(3+\sqrt{8}{)}^2}$

$\Rightarrow 17+6\sqrt{18}+\frac{1}{17+6\sqrt{8}}$

$\Rightarrow \frac{(17+6\sqrt{8}{)}^2+1}{17+6\sqrt{8}}\times \frac{(17-6\sqrt{8})}{(17-6\sqrt{8})}$

$\Rightarrow \frac{(17+6\sqrt{18}{)}^2(17-6\sqrt{8})+(17-6\sqrt{8})}{289-288}$

$\Rightarrow (289-288)(17+6\sqrt{8})+17-6\sqrt{8}$

$\Rightarrow 17+6\sqrt{8}+17-6\sqrt{8}=34$